# Find the longest sub-string which is prefix, suffix and also present inside the string

Given a string str. The task is to find the longest sub-string which is a prefix, a suffix and a sub-string of the given string, str. If no such string exists then print -1.

Examples:

Input: str = “fixprefixsuffix”
Output: fix
“fix” is a prefix, suffix and present inside in the string too.

Input: str = “aaaa”
“aa” is a prefix, suffix and present inside the string.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Let us calculate the longest prefix suffix for all prefixes of string. longest prefix suffix lps[i] is maximal length of prefix that also is suffix of substring [0…i]. More about longest prefix suffix you can see in a description of kmp algorithm.

The first of possible answers is a prefix of length lps[n-1]. If lps[n-1] = 0, there is no solution. For checking the first possible answer you should iterate over lps[i]. If at least one of them equal to lps[n-1] (but not n-1th, of course) – you found the answer. The second possible answer is a prefix of length lps[lps[n-1]-1]. If lps[lps[n-1]-1] = 0, you also have no solution. Otherwise, you can be sure that the answer already found. This substring is a prefix and a suffix of our string. Also, it is a suffix of a prefix with length lps[n-1] that places inside of all string. This solution works in O(n).

Below is the implementation of the above approach:

 // C++ implementation of the approach #include using namespace std;    // Function to find longest prefix suffix vector compute_lps(string s) {     int n = s.size();        // To store longest prefix suffix     vector lps(n);        // Length of the previous     // longest prefix suffix     int len = 0;        // lps[0] is always 0     lps[0] = 0;     int i = 1;        // Loop calculates lps[i] for i = 1 to n - 1     while (i < n) {         if (s[i] == s[len]) {             len++;             lps[i] = len;             i++;         }            // (pat[i] != pat[len])         else {             if (len != 0)                 len = lps[len - 1];             // Also, note that we do not increment             // i here                // If len = 0             else {                 lps[i] = 0;                 i++;             }         }     }        return lps; }    // Function to find the longest substring // which is prefix as well as a // sub-string of s[1...n-2] void Longestsubstring(string s) {     // Find longest prefix suffix     vector lps = compute_lps(s);     int n = s.size();        // If lps of n-1 is zero     if (lps[n - 1] == 0) {         cout << -1;         return;     }        for (int i = 0; i < n - 1; i++) {            // At any position lps[i] equals to lps[n - 1]         if (lps[i] == lps[n - 1]) {             cout << s.substr(0, lps[i]);             return;         }     }        // If answer is not possible     if (lps[lps[n - 1] - 1] == 0)         cout << -1;     else         cout << s.substr(0, lps[lps[n - 1] - 1]); }    // Driver code int main() {     string s = "fixprefixsuffix";        // function call     Longestsubstring(s);        return 0; }

 // Java implementation of the approach class GFG {     // Function to find longest prefix suffix     static int [] compute_lps(String s)     {         int n = s.length();                // To store longest prefix suffix         int [] lps = new int [n];                // Length of the previous         // longest prefix suffix         int len = 0;                // lps[0] is always 0         lps[0] = 0;         int i = 1;                // Loop calculates lps[i] for i = 1 to n - 1         while (i < n)          {             if (s.charAt(i) == s.charAt(len))              {                 len++;                 lps[i] = len;                 i++;             }                    // (pat[i] != pat[len])             else              {                 if (len != 0)                     len = lps[len - 1];                 // Also, note that we do not increment                 // i here                        // If len = 0                 else                 {                     lps[i] = 0;                     i++;                 }             }         }                return lps;     }            // Function to find the longest substring     // which is prefix as well as a     // sub-string of s[1...n-2]     static void Longestsubstring(String s)     {         // Find longest prefix suffix         int [] lps = compute_lps(s);         int n = s.length();                // If lps of n-1 is zero         if (lps[n - 1] == 0)         {             System.out.println(-1);             return;         }                for (int i = 0; i < n - 1; i++)          {                    // At any position lps[i] equals to lps[n - 1]             if (lps[i] == lps[n - 1])              {                 System.out.println(s.substring(0, lps[i]));                 return;             }         }                // If answer is not possible         if (lps[lps[n - 1] - 1] == 0)             System.out.println(-1);         else             System.out.println(s.substring(0, lps[lps[n - 1] - 1]));     }            // Driver code     public static void main (String [] args)     {         String s = "fixprefixsuffix";                // function call         Longestsubstring(s);            } }    // This code is contributed by ihritik

 # Python3 implementation of the approach    # Function to find longest prefix suffix def compute_lps(s):        n = len(s)        # To store longest prefix suffix     lps = [0 for i in range(n)]        # Length of the previous     # longest prefix suffix     Len = 0        # lps[0] is always 0     lps[0] = 0     i = 1        # Loop calculates lps[i] for i = 1 to n - 1     while (i < n):         if (s[i] == s[Len]):             Len += 1             lps[i] = Len             i += 1            # (pat[i] != pat[Len])         else:             if (Len != 0):                 Len = lps[Len - 1]             # Also, note that we do not increment             # i here                # If Len = 0             else:                 lps[i] = 0                 i += 1                       return lps    # Function to find the longest substring # which is prefix as well as a # sub-of s[1...n-2] def Longestsubstring(s):        # Find longest prefix suffix     lps = compute_lps(s)     n = len(s)        # If lps of n-1 is zero     if (lps[n - 1] == 0):         print(-1)         exit()            for i in range(0,n - 1):            # At any position lps[i] equals to lps[n - 1]         if (lps[i] == lps[n - 1]):             print(s[0:lps[i]])             exit()        # If answer is not possible     if (lps[lps[n - 1] - 1] == 0):         print(-1)     else:         print(s[0:lps[lps[n - 1] - 1]])    # Driver code    s = "fixprefixsuffix"    # function call Longestsubstring(s)    # This code is contributed by mohit kumar

 // C# implementation of the approach using System;    class GFG {     // Function to find longest prefix suffix     static int [] compute_lps(string s)     {         int n = s.Length;                // To store longest prefix suffix         int [] lps = new int [n];                // Length of the previous         // longest prefix suffix         int len = 0;                // lps[0] is always 0         lps[0] = 0;         int i = 1;                // Loop calculates lps[i] for i = 1 to n - 1         while (i < n)          {             if (s[i] == s[len])             {                 len++;                 lps[i] = len;                 i++;             }                    // (pat[i] != pat[len])             else              {                 if (len != 0)                     len = lps[len - 1];                 // Also, note that we do not increment                 // i here                        // If len = 0                 else                  {                     lps[i] = 0;                     i++;                 }             }         }                return lps;     }            // Function to find the longest substring     // which is prefix as well as a     // sub-string of s[1...n-2]     static void Longestsubstring(string s)     {         // Find longest prefix suffix         int [] lps = compute_lps(s);         int n = s.Length;                // If lps of n-1 is zero         if (lps[n - 1] == 0)          {             Console.WriteLine(-1);             return;         }                for (int i = 0; i < n - 1; i++)         {                    // At any position lps[i] equals to lps[n - 1]             if (lps[i] == lps[n - 1])              {                 Console.WriteLine(s.Substring(0, lps[i]));                 return;             }         }                // If answer is not possible         if (lps[lps[n - 1] - 1] == 0)             Console.WriteLine(-1);         else             Console.WriteLine(s.Substring(0, lps[lps[n - 1] - 1]));     }            // Driver code     public static void Main ()     {         string s = "fixprefixsuffix";                // function call         Longestsubstring(s);            } }    // This code is contributed by ihritik



Output:
fix

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